# Circular Motion with Decreasing Radius

• uSee2
In summary, the conversation discusses the relationship between tangential speed and radius in a circular motion scenario. The proof shows that the centripetal force is equal to the tangential force, and when the radius is halved, the tangential speed must also be halved in order to maintain this equality. The attempt to double the tangential speed by decreasing the radius and increasing the speed is proven to be incorrect. The correct relationship is that the new tangential speed is half the original speed when the radius is halved.

#### uSee2

Homework Statement
One end of a string is attached to a ball, with the other end held by a student such that the ball is swung in a horizontal circular path of radius R at a constant tangential speed. At a later time, the tension force exerted on the ball remains constant, but the length of the string is decreased to R/4. What is the new tangential speed of the ball?
Relevant Equations
##F_c = \frac {mv^2} R##
The answer key states that the new tangential speed is half the original speed. However, this isn't correct right? It should double.

My proof:

##F_c = \frac {mv^2} R##
##F_c = F_t##
##\frac {mv^2} {\frac R 4} = \frac {m(2v)^2} R## If centripetal force were to stay constant.
As such, tangential velocity should double right? It should not half?

Your logic is faulty. The centripetal force is, as you say, ##F_c =\frac{mv^2}{R}.## For the fraction to remain constant, if the denominator is decreased by a factor of 4, the numerator must also decrease by a factor of 4. Since the speed in the numerator is squared (##v^2=v*v##), each of the v's must decrease by a factor of 2 when the radius is decreased by a factor of 4.

uSee2, scottdave and topsquark
kuruman said:
Your logic is faulty. The centripetal force is, as you say, ##F_c =\frac{mv^2}{R}.## For the fraction to remain constant, if the denominator is decreased by a factor of 4, the numerator must also decrease by a factor of 4. Since the speed in the numerator is squared (##v^2=v*v##) each of the v's must decrease by a factor of 2 when the radius is decreased by a factor of 4.
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?

uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?
Try using two different symbols for the first velocity and the second velocity.

$$\frac {mv_1^2} {R} = \frac {mv_2^2} {\frac R 4}$$

Well I cannot get my Latex to display correctly, but try it on your own.

Last edited:
uSee2 and TSny
uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?
No. The new ratio must be the same as the old ratio.
Old ratio = ##\frac{mv^2}{R}##.
New ratio (according to you) = ##\frac{m(2v)^2}{R}.##

Are they the same?

uSee2
uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?
Yes, and that is exactly what the answer key says. The new tangential speed, v, is half the original, 2v.

uSee2
haruspex said:
Yes, and that is exactly what the answer key says. The new tangential speed, v, is half the original, 2v.
kuruman said:
No. The new ratio must be the same as the old ratio.
Old ratio = ##\frac{mv^2}{R}##.
New ratio (according to you) = ##\frac{m(2v)^2}{R}.##

Are they the same?
I do see that they aren't the same. But when I tried it again with different symbols just like @scottdave said:

2 is after the label for when radius is reduced, 1 is the label for before.

##F_{c2} = \frac {mv_2^2} {R_2}##

##F_{c1} = \frac {mv_1^2} {R_1}##

##F_{c1} = F_{c2}##

##R_2 = \frac 1 4 {R_1}##

##\frac {mv_1^2} {R_1}= \frac {mv_2^2} {R_2}##

##\frac {mv_1^2} {R_1}= \frac {mv_2^2} {\frac 1 4 R_1}## Substitution of ##R_2 = \frac 1 4 {R_1}##

Multiply each side by ##\frac {R_1} {m}##

##\frac {v_2^2} {\frac 1 4} = v_1^2##

##v_2^2 = 4v_1^2##

Take the square root of each side

##v_2 = 2v_1##

The new speed is double the original speed I think, I may have made an error though.

uSee2 said:
##\frac {v_2^2} {\frac 1 4} = v_1^2##

##v_2^2 = 4v_1^2##
Check how you went from the first equation to the second equation.

scottdave and uSee2
TSny said:
Check how you went from the first equation to the second equation.
Ohh!! I can't believe I made that mistake. Thank you so much for catching that. I see now that ##v_2 = \frac 1 2 v_1##

scottdave